Download pure liquid-vapour equilibrium - Theoretical and Computational

CHEM 2PA3, 2004
Experiment 3
PURE LIQUID-VAPOUR EQUILIBRIUM
OBJECTIVE
Determine the vapour pressure of a liquid as a function of temperature by the isoteniscope method, and
calculate the enthalpy of vapourisation of the liquid.
INTRODUCTION
If the pressure on a gas/vapour is increased at constant temperature its volume will decrease (isotherm
T3 in Fig.1). At temperatures below TC e.g. T1, when the pressure reaches a definite value the volume is
observed to decrease markedly for an infinitesimal increase in pressure (A-B in Fig. 1). The marked
decrease in volume is due to the formation of increasing amounts of liquid, commencing at A, until at
point B the gas/vapour is completely converted to liquid. The line A-B represents equilibrium between
the pure liquid and its vapour.
Fig.1. Isotherms in the region of liquid vapour equilibrium.
The constant pressure P1 is the equilibrium vapour pressure of the liquid at temperature T1. As the
temperature is increased, the vapour pressure of the liquid increases also. The temperature Tb for
which the liquid has a vapour pressure exactly equal to 1 atmosphere is called the normal boiling point
of the liquid. At higher temperatures such as T2 the horizontal portions of the isotherms become
shorter. For example the length C - D is shorter than A - B. This is because the pressure P2 > P1 and
the molar volumes of the vapour become smaller.
CHEM 2PA3, 2004
Experiment 3
The temperature at which the horizontal portion just vanishes is called the CRITICAL TEMPERATURE,
Tc, and the corresponding (vapour) pressure is called the CRITICAL PRESSURE, Pc. The shaded area
in Fig.1 includes all states of the system which correspond to the existence of two phases, liquid and
vapour.
Part of the shaded area, up to the normal boiling point, Tb, of the liquid, is the subject of this experiment
in which the vapour pressure dependence on temperature will be investigated. This dependence can be
related to the enthalpy of vapourisation Hvap in the following way. For equilibrium between the liquid
and vapour at constant temperature and pressure, the liquid and vapour must be at the same chemical
potential.
[1]
where for a one-component system
= G , the molar Gibbs free energy.
If this relationship is to be satisfied over the whole temperature range for which liquid and vapour can
co-exist it must follow that T and P are not independent variables. For an infinitesimal change in
temperature, if equilibrium is to be maintained
[2]
Since
[3]
we can write
[4]
and
[5]
where ∆S vap and ∆Vvap are the entropy and volume changes accompanying the transfer of one mole
of liquid to the vapour state at constant P and T. At equilibrium,
[6]
and [5] becomes
[7]
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CHEM 2PA3, 2004
Experiment 3
This relationship [7] is known as the CLAPEYRON EQUATION. It may be simplified by making the
approximation Vvap = Vv. The molar volume of the liquid can be neglected compared to that of the
vapour. Defining the compressibility factor of the vapour as
[8]
we can write
[9]
For an ideal gas z = 1 and we obtain the Clausius-Clapeyron equation
[10]
Thus if the vapour were indeed an ideal gas and
limits of the approximation
H vap independent of temperature then, within the
Vvap = Vv, a plot of lnP against 1/T would be a straight line of slope - H vap
/R. Since the plot is quite linear for most liquids we can deduce that, at least,
constant. Note that
H vap /z is fairly
H vap and z are both temperature dependent parameters.
[11]
where
C p = C p (vapour) – C p (liquid). z can be estimated from the universal charts based on the
law of corresponding states or by use of an equation of state such as the Berthelot equation.
[12]
where Pr and Tr are the reduced pressure and temperature.
[NB. Pr = P/Pc and Tr = T/Tc where Pc and Tc are the critical pressure and temperature.]
PROCEDURE
The isoteniscopic method will be used to complete this lab. A diagram of the apparatus is shown in
Figure 2 below. The principle of the method is that when the liquid levels inside the U-tube of the
isoteniscope are equal, the vapour pressure of the liquid in the bulb acting on the one side of the Utube is equal to the external pressure acting on the other side of the U-tube. The latter is measured with
the closed-end mercury manometer. Do not forget to record the room temperature and pressure
located at the last fume hood in room ABB 107.
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CHEM 2PA3, 2004
Experiment 3
Fig. 2. Vapour-Pressure Apparatus – The Isoteniscope Method
1.
Assemble the apparatus as shown in Fig.2 but do not connect the isoteniscope.
2.
Prepare enough half-and-half slurry of crushed ice and water to fill the water bath. Set the
temperature control on the heater/stirrer unit for stirring only.
3.
Fill the isoteniscope bulb approximately 3/4 full with the liquid under investigation and fill the Utube with the same liquid to about half-way up the enlarged section of the isoteniscope.
a. This may be a little trying, be sure to ask your TA for help if you need it.
4.
Connect the isoteniscope to the remainder of the apparatus.
5.
Start the stirrer and turn the stopcock S to the vacuum pump and evacuate air trapped between
the isoteniscope bulb and the U-tube. It is worthwhile checking the first point against a published
value for the vapour pressure [2] to ensure that all the air has been removed. Otherwise the
partial pressure of the residual air may make your subsequent work useless.
6.
Equalize the levels of the liquid in the U-tube taking great care not to re-admit air into the
isoteniscope bulb. Should this happen, then you must remove it before proceeding further.
7.
Allow time for the liquid to assume the temperature of the bath, note its temperature and read
the mercury manometer.
8.
Raise the temperature of the bath by about 5 °C and repeat the measurements of temperature
and pressure.
9.
Continue raising the temperature of the bath by 5 °C until no further measurement of pressure is
possible.
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CHEM 2PA3, 2004
Experiment 3
CALCULATIONS
Record room temperature and correct the measured mercury columns to equivalent ones at 0 °C using
the relation
[13]
where Pm is the measured pressure and Pa is the corrected total pressure to be used in computations
and t is room temperature in °C. Plot the logarithm of the vapour pressure against the reciprocal of the
absolute temperature. Determine the slope, A, and intercept, B, of this plot by two methods.
[14]
1.
Draw the best straight line through the points by eye and determine the values of A and B from
this line.
2.
Review the theory of Least-Squares Analysis to calculate A and B for this line. You are required
to perform this analysis with any spreadsheet. Calculate Hvap from the visually estimated and
least-squares determined slopes using [10].
This value for the H vap is inaccurate because of the assumptions made in deriving the ClausiusClapeyron equation. Determine the compressibility factor for a particular point, say 25 °C, on the ln P vs
1/T curve and use it together with the slope of the curve at that point to calculate the corrected H vap .
Deduce from the graph the normal boiling point of the liquid and compare it with the accepted value.
DISCUSSION
Discuss critically the experimental procedure paying attention to accuracy. Compare H vap values
obtained from the visual slope, the least-squares analysis and the literature. For most liquids the plots
of ln P versus (1/T) and equation (10) predicts the decrease of H vap with increase in temperature.
Account for this apparent paradox. Could the liquid used be classified as "normal" according to
Trouton's Rule?
REFERENCES
1. Shoemaker, Garland, Steinfeld and Nibler, "Experiments in Physical Chemistry", 4th edition,
McGraw-Hill, pp. 197 ff.
2. Any CRC Handbook.
3. J.H. Noggle, "Physical Chemistry", 3rd ed., Harper Collins, 1996, pp 175 - 180.
4. R.G. Mortimer "Physical Chemistry", Benjamin/Cummings, Redwood City, Calif., 1993, 176-183.
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